Answer
a. $ f(2\pi/3)=\frac{\pi}{3}-\sqrt 3$ local minimum, $f(0)=0$ and $f(2\pi)=\pi$ local maxima.
$ f(2\pi/3)=\frac{\pi}{3}-\sqrt 3$ absolute minimum and $f(2\pi)=\pi$ absolute maximum.
b. See graph and explanations.
Work Step by Step
a. Step 1. Given the function $f(x)=\frac{x}{2}-2sin(\frac{x}{2}), 0\leq x\lt2\pi$, we have $f'(x)=\frac{1}{2}-cos(\frac{x}{2})$.
Step 2. The extrema happen when $f'(x)=0$, undefined, or at endpoints; thus, we have $cos(\frac{x}{2})=\frac{1}{2}$ and the critical points within the domain are $x=0, 2\pi/3, 2\pi$.
Step 3. We have $f(0)=0, f(2\pi/3)=\frac{\pi}{3}-2sin(\frac{\pi}{3})=\frac{\pi}{3}-\sqrt 3, f(2\pi)=\pi$. Test signs of $f'(x)$ across critical points $(0)...(-)...(2\pi/3)..(+)..(2\pi)$; thus, we have $ f(2\pi/3)=\frac{\pi}{3}-\sqrt 3$ as a local minimum, $f(0)=0$ and $f(2\pi)=\pi$ as local maxima.
Step 4. We can identify $ f(2\pi/3)=\frac{\pi}{3}-\sqrt 3$ as an absolute minimum and $f(2\pi)=\pi$ as an absolute maximum in the interval.
b. See graph. We can see that when $f'\lt0$, $f$ decreases; while when $f'\gt0$, $f$ increases (both with the tangent slopes corresponding to the values of $f'$). When $f'=0$, $f$ will be at an extrema with tangent slopes of zero. We can also observe the change of signs in $f'$ on different sides of critical points where $f'=0$.
